
According to Amitsur–Levitsky theorem \(s_4\)  evaluated on \(2\times 2\)  matrices is a PI, and thus is PI on quaternions, as is its multilinearization.

The Lie bracket \(p(x,y)=xy-yx\)  is a Lie polynomial and its image is \(V\) .

The polynomial \([x_1,x_2][x_3,x_4]+[x_3,x_4][x_1,x_2]\) , the multilinearization of \([x_1,x_2]^2\) , evaluated on \(\mathbb {H}\) , is a multilinear central polynomial.

An example where \(\operatorname{Im}p=\mathbb {H}(\mathbb {R})\)  can be constructed trivially.

The central polynomial \(p(x,y)=[x,y]^4\)  provides an example of complete homogeneous polynomial evaluated of \(2\times 2\)  matrices with real entries and taking only positive central values I took a square of \(p\) . Note that evaluated on quaternions we do not need to take a square. This polynomial takes only non positive values: indeed, \((ai+bj+ck)^2=-a^2-b^2-c^2\) , and \(\operatorname{Im}p\)  is \(\mathbb {R}_{\le 0}\) . Hence, \(-p(x,y)=-[x,y]^2\)  is an example of the polynomial with image set \(\mathbb {R}_{\ge 0}\) .

The polynomial \(p(x,y)=[x,y]^2+\big [x^2,y^2\big ]\)  is the sum of two polynomials, the image of the first term is \(\mathbb {R}_{\le 0}\) , and the second one has image \(V\) ,
thus \(\operatorname{Im}p\)  is the set of quaternions with non-positive real part. Of course \(-p\)  has the opposite image: the set of quaternions with non-negative real part.

In Example REF (i), the polynomial \(g(x_1 , x_2 ) = [x_1 , x_2 ]^2\)  has the property that \(g(A, B) = 0\) 
whenever \(A\)  is scalar, but \(g\)  can take on a non-zero value whenever \(A\)  is non-scalar. Thus, \(g(x_1 , x_2 )x_1\)  takes all values except scalars.

Any quaternion \(q=a+v\) , \(a\in \mathbb {R}\) , \(v\in V\)  can be considered as a \(2\times 2\)  matrix with complex entries, whose eigenvalues are \(\lambda _{1,2}=a\pm ni\) , where \(n=\left|\left|v\right|\right|\)  is the norm of the vector part of the quaternion. In particular
\(\operatorname{tr}q=2a=2\operatorname{Re}q\)  and \(\det q=a^2+n^2=\left|\left|q\right|\right|\) . In Example REF (ii) we provided an example of the polynomial taking all possible values except for those where the ratio of eigenvalues belongs to some set \(S\) . However here we should have a polynomial with real coefficients. Nevertheless this is possible:
let \(S\)  be any finite subset of \(S^2\)  – a unit circle on the complex plane. There exists a
completely homogeneous polynomial \(p\)  such that \(\operatorname{Im}p\)  is the
set of all quaternions except the quaternions with ratio of
eigenvalues from \(S\) . Any \(c\in S\)  can be written in the form \(c=\frac{a+bi}{a-bi}\)  for some \(a,b\in \mathbb {R}\) . The construction of the polynomial is as follows. Consider
\(f(x)=x\cdot \prod _{c\in S}((a+bi)\lambda _1-(a-bi)\lambda _2)((a+bi)\lambda _2-(a-bi)\lambda _1),\) 
where
\(\lambda _{1,2}\)  are eigenvalues of \(x\)  and \(c=\frac{a+bi}{a-bi}\) . For each \(c\)  the
product \((a+bi)\lambda _1-(a-bi)\lambda _2)((a+bi)\lambda _2-(a-bi)\lambda _1)=-a^2(\lambda _1-\lambda _2)^2-b^2(\lambda _1+\lambda _2)^2\) 
is
a polynomial with real coefficients in \(\operatorname{tr}x\)  and \(\operatorname{tr}x^2\) . Thus
\(f(x)\)  is a polynomial with traces, and by [1]}, one can rewrite each trace in \(f\)  as a fraction of
multilinear central polynomials.
After that we multiply the expression by the product of all the
denominators, which we can take to have value 1. We obtain a
completely homogeneous polynomial \(p\)  which image is the cone
under \(\operatorname{Im}f\)  and thus equals \(\operatorname{Im}f\) . The image of \(p\)  is
the set of all quaternions with ratios of eigenvalues
not belonging to \(S\) .


